What is a tight lower bound to $\sum _{i=1}^{n}\frac{1}{a+{x}_{i}}$ under the restrictions $\sum _{i=1}^{n}{x}_{i}=0$ and $\sum _{i=1}^{n}{x}_{i}^{2}={a}^{2}$ ?

Conjecture: due to the steeper rise of $\frac{1}{a+x}$ for negative $x$, one may keep those values as small as possible. So take $n-1$ values ${x}_{i}=-q$ and ${x}_{n}=(n-1)q$ to compensate for the first condition. The second one then gives ${a}^{2}=\sum _{i=1}^{n}{x}_{i}^{2}={q}^{2}((n-1{)}^{2}+n-1)={q}^{2}n(n-1)$. Hence,

$\sum _{i=1}^{n}\frac{1}{a+{x}_{i}}\ge \frac{n-1}{a(1-1/\sqrt{n(n-1)})}+\frac{1}{a(1+(n-1)/\sqrt{n(n-1)})}$

should be the tight lower bound.

Conjecture: due to the steeper rise of $\frac{1}{a+x}$ for negative $x$, one may keep those values as small as possible. So take $n-1$ values ${x}_{i}=-q$ and ${x}_{n}=(n-1)q$ to compensate for the first condition. The second one then gives ${a}^{2}=\sum _{i=1}^{n}{x}_{i}^{2}={q}^{2}((n-1{)}^{2}+n-1)={q}^{2}n(n-1)$. Hence,

$\sum _{i=1}^{n}\frac{1}{a+{x}_{i}}\ge \frac{n-1}{a(1-1/\sqrt{n(n-1)})}+\frac{1}{a(1+(n-1)/\sqrt{n(n-1)})}$

should be the tight lower bound.